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### equilibrium review

Posted: Fri Mar 13, 2020 8:50 am
Could someone help me with number 4 on the UA first review sheet? I am a bit confused as to where to even start with this problem.

Q4. The overall dissociation of oxalic acid, H2C2O4
is represented below. The overall
dissociation constant is also indicated. a) What volume of 0.400 M NaOH is required to
neutralize completely a 5.00 x 10-3
mol sample of pure oxalic acid. b) Give the equations
representing the first and second dissociations of oxalic acid. Calculate the value of the first
dissociation constant, K1
, for oxalic acid if the value of the second dissociation constant, K2
, is 6.40 x 10-5
. c) A strong acid is added to a 0.015 M solution of oxalic acid until the pH is 0.5.
Calculate the [C2O42-] in the resulting solution. (Assume the change in volume is negligible.)
H2C2O4 ⇌ 2H+ + C2O42-
, K = 3.78 x 10-6

### Re: equilibrium review

Posted: Fri Mar 13, 2020 1:11 pm
Sarah Blake-2I wrote:Could someone help me with number 4 on the UA first review sheet? I am a bit confused as to where to even start with this problem.

Q4. The overall dissociation of oxalic acid, H2C2O4
is represented below. The overall
dissociation constant is also indicated. a) What volume of 0.400 M NaOH is required to
neutralize completely a 5.00 x 10-3
mol sample of pure oxalic acid. b) Give the equations
representing the first and second dissociations of oxalic acid. Calculate the value of the first
dissociation constant, K1
, for oxalic acid if the value of the second dissociation constant, K2
, is 6.40 x 10-5
. c) A strong acid is added to a 0.015 M solution of oxalic acid until the pH is 0.5.
Calculate the [C2O42-] in the resulting solution. (Assume the change in volume is negligible.)
H2C2O4 ⇌ 2H+ + C2O42-
, K = 3.78 x 10-6

Well first write a balanced equation for the neutralization of oxalic acid and sodium hydroxide with the dissociation of a single H first, and then do the ICE table.